Golden Ratio In a 2x2 Square: Without And Within

Assume $AO=OB=BC=OD\;$ and $OD\perp AB.\;$ $M\;$ is the intersection of $AB\;$ with the circle $C(D),\;$ centered at $C\;$ and passing through $D.$

Then $M\;$ divides $AB\;$ in the Golden Ratio: $\displaystyle\frac{BM}{AM}=\varphi.$

The proof is also straightforward: assume $AO=OB=BC=OD=1.\;$ Then $CM=CD=\sqrt{5};\;$ $BM=\sqrt{5}-1;\;$ $AM=3-\sqrt{5},\;$ so that $\displaystyle\frac{BM}{AM}=\frac{\sqrt{5}-1}{3-\sqrt{5}}=\frac{\sqrt{5}+1}{2}=\varphi.$

Dung Thanh Nguyen's construction can be represented a little differently, starting with a $2\times 2\;$ square:

This construction is responsible to the Without part in the title of the page. The Within part is illustrated below.

In all three diagrams the ratio of the red segment to the blue one is Golden.